LING/C SC/PSYC 438/538 Computational Linguistics Sandiway Fong Lecture 8: 9/18

Administrivia Last Thursday: –class replaced by –Cognitive Science Master’s Seminar Series talk on Statistical Natural Language Parsing and Treebanks

Administrivia This Thursday –class replaced by this Friday’s Linguistics Colloquium

Administrivia This Friday: –Computational Linguistics Colloquium –Presenters: Sandiway Fong, Mike Hammond and Ying Lin –Time: 3:00-4:30 PM –Location: Speech and Hearing 205H –Abstract: –In this talk, we will demonstrate the use of various software to collect and analyze linguistic data of various sorts. –Fong will give a tutorial on using his freely-available treebankviewer software to explore syntactic structures in treebanks. –Hammond will demonstrate the use of his "Finite State Playground", a suite of command-line tools that can be used to construct and manipulate finite-state language models. –Lin will discuss the use of speech technology and statistical tools for phonetic data analysis.

Today’s Topics DCG topics –language enumeration –ambiguity: multiple parses –stepping outside regular grammars just a little bit Homework 2

Regular Grammars: Language Enumeration right recursive regular grammar: s --> [b], ays. ays --> [a], a. a --> [a], exclam. a --> [a], a. exclam --> [!]. language: Sheeptalk –baa! –baaa! –ba...a! this DCG always halts: ?- s([b,a,a,!],[]). Yes ?- s([b,a,!],[]). No as a decider But not in all modes of operation –Prolog’s top-down, left-to-right, depth-first strategy also allows the grammar to be used an enumerator ?- s(L,[]). –L is a variable, we’re not naming the list. Let Prolog figure it out.

Regular Grammars: Language Enumeration Try ?- s(L,[]). –Prolog returns one answer and waits for action (type ; to get the next, to finish) Output: ?- s(L,[]). L = [b,a,a,!] ? ; L = [b,a,a,a,!] ? ; L = [b,a,a,a,a,!] ? ; L = [b,a,a,a,a,a,!] ? ; L = [b,a,a,a,a,a,a,!] ? ; L = [b,a,a,a,a,a,a,a,!] ? ; L = [b,a,a,a,a,a,a,a,a,!] ? Prolog is enumerating the strings of the language

Regular Grammars: Language Enumeration Let’s see why this happens ?- s(L,[]). DCG : 1.s --> [b], ays. 2.ays --> [a], a. 3.a --> [a], exclam. 4.a --> [a], a. 5.exclam --> [!]. derivation tree: s bays aa a exclam ! bays aa a a s aexclam !

Regular Grammars: Language Enumeration left recursive regular grammar: s --> a, [!]. a --> firsta, [a]. a --> a, [a]. firsta --> b, [a]. b --> [b]. doesn’t halt when faced with a string not in the language Example: ?- s([b,a,!],[]). enters an infinite loop Question: –how well does it do when enumerating? –i.e. ?- s(L,[]). –does this query halt?

Regular Grammars: Language Enumeration left recursive regular grammar: 1.s --> a, [!]. 2.a --> firsta, [a]. 3.a --> a, [a]. 4.firsta --> b, [a]. 5.b --> [b]. doesn’t halt when faced with a string not in the language Perhaps surprisingly: ?- s(L,[]). enumerates just fine. derivation tree: s a firsta b b a a ! s a a b b a a a ! and so on

Regular Grammars: Language Enumeration However, this slightly re-ordered left recursive regular grammar: 1.s --> a, [!]. 2.a --> a, [a]. 3.a --> firsta, [a]. 4.firsta --> b, [a]. 5.b --> [b]. won’t halt when enumerating Why? s a a a a... descends infinitely using rule 2

DCG: Ambiguity A context-free grammar that returns a parse: avoiding some nasty recursion for infinite loops s(s(NP,VP)) --> np(NP), vp(VP). np(np(i)) --> [i]. np(np(D,N)) --> det(D), noun(N). np(np(D,N,PP)) --> det(D), noun(N), pp(PP). det(d(the)) --> [the]. det(d(a)) --> [a]. noun(n(boy)) --> [boy]. noun(n(telescope)) --> [telescope]. vp(vp(V,NP)) --> verb(V), np(NP). vp(vp(V,NP,PP)) --> verb(V), np(NP), pp(PP). verb(v(saw)) --> [saw]. pp(pp(P,NP)) --> p(P), np(NP). p(p(with)) --> [with].

DCG: Ambiguity Parsing : ?- s(X,[i,saw,the,boy,with,a,telescope],[]). X = s(np(i),vp(v(saw),np(d(the),n(boy),pp(p(with),np(d(a),n(telescope)))))) ? ; X = s(np(i),vp(v(saw),np(d(the),n(boy)),pp(p(with),np(d(a),n(telescope))))) ? ; no Prolog finds both parses by backtracking. But why does Prolog return the parses in this order? 5min exercise: re-arrange grammar to produce the opposite order of parses

Regular Grammars: Stepping Outside Beyond regular grammars –a n b n = {ab, aabb, aaabbb, aaaabbbb,... } n>=1 is not a regular language (accept fact for now) A regular grammar extended to allow both left and right recursive rules can accept/generate it a --> [a], b. b --> [b]. b --> a, [b]. B A b a A B A b a Example: this grammar accepts –aabb aaaabbbb and rejects –aab aaaabbbbb Intuition: –grammar implements the stacking of partial trees balanced for a’s and b’s:

Homework 2 Ground rules: –all in one single file please –no more multiple attachments

Homework 2 Question 1 –Consider a language L 3or5 = {111, 11111, , , , , ,...} –each member of L 3or5 is a string containing only 1s –the number of 1s in each string is divisible by 3 or 5 Give a regular grammar that generates language L 3or5 Show your Prolog code works on strings in the language and rejects strings outside the language

Homework Question 2: Language L = {a 2n b n+1 | n>=1} is also non-regular nearly twice as many a’s as in the b’s but there’s always one b too many for this to be true but can be generated by a regular grammar extended to allow left and right recursive rules Given a Prolog grammar satisfying these rules for L Show it accepts and rejects correctly Legit rules: X -> aY X -> a X -> Ya –where X, Y are non- terminals, a is some arbitirary terminal

Homework 2 Question 3 (Optional 438) A right recursive regular grammar that generates a (rightmost) parse for the language {a n | n>=2}: s(s(a,B)) --> [a], b(B). b(b(a,B)) --> [a], b(B). b(b(a))--> [a]. Example ?- s(Tree,[a,a,a],[]). Tree = s(a,b(a,b(a)))? A corresponding left recursive regular grammar will not halt in all cases when an input string is supplied Modify the right recursive grammar to produce a left recursive parse, e.g. ?- s(Tree,[a,a,a],[]). Tree = s(b(b(a),a),a) Can you do this for any right recursive regular grammar? e.g. the one for sheeptalk Explain your answer